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Formula for sum of digits of a number

  1. May 19, 2012 #1
    Any ideas on this? I googled it and got only answers inolving Excel, C++, etc...
    A formula for finding the digit sum of a 2-digit number, or only a 3-digit number, would also be interesting.

    I thought I had a start with this: For a 2-digit number xy, the sum of the digits = [itex](\frac{x}{10})\lfloor[/itex] + something. But I don't know how to isolate y.
     
  2. jcsd
  3. May 19, 2012 #2

    chiro

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    Hey greg9381 and welcome to the forums.

    The best way to answer this for any base is to use the DIV\MOD algorithm. This will generate each digit for each base-position for any base and then you can simply sum all the base-position components and get an answer.

    You can apply this say for a number in base 2, base 16, or base 10. From the sounds of it, you want to apply this algorithm to base 10. Also I'm going to assume you have a whole number: if you don't just remove the remainder (term after the decimal point) and treat this result as a normal whole number.

    Take a look at this:

    http://www.cut-the-knot.org/recurrence/conversion.shtml
     
  4. May 19, 2012 #3
    Well, I'm not very advanced (AB calc in high school) so I can't completely understand the article, but it seems to be a way to find the digit sum using a computer? To clarify, I'm wondering if it's possible to have a formula that can be done with pencil and paper, not computers.
     
  5. May 20, 2012 #4

    chiro

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    The basic idea is very simple and goes a little something like this:

    Take a number n. Do the following until you get n = 0.

    Set i = 1
    Calculate a(i) = n MOD 10. Let n = n - a(i). n = n / 10.
    If n = 0 then we are done. If not then increase i by 1 and go back to the previous step.

    Then if you want the sum simply add up all the a(i)'s and that will give you your sum.

    This is for base-10, but if you want to do it for any base then just replace 10 with the base you are working with.

    Quick example. Let n = 231.

    a(1) = 231 MOD 10 = 1. 231 - 1 = 230. 230/10 = 23 which is not zero so i = 2.
    a(2) = 23 MOD 10 = 3. 23 - 3 = 20. 20/10 = 2 which is not zero so i = 3.
    a(3) = 2 MOD 10 = 2, 2-2 = 0, 0/10 = 0 so we stop.

    Now add up all a(i)'s to get a(1) + a(2) +a(3) = 1 + 3 + 2 = 5 which is what we expect.
     
  6. Mar 20, 2013 #5
    I got it.

    About thirty minutes of non-continuous free time and a half sheet of paper gave me this:
    For any whole number "x":
    1 digit: x
    2 digits: ([itex]\left\lfloor[/itex][itex]\frac{x}{10}[/itex][itex]\right\rfloor[/itex]) + (x -10([itex]\left\lfloor[/itex][itex]\frac{x}{10}[/itex][itex]\right\rfloor[/itex]))
    ...

    After awhile, I realized I could combine like terms, and this is what I got:

    x - 9([itex]\left\lfloor[/itex][itex]\frac{x}{10}[/itex][itex]\right\rfloor[/itex]) - 9([itex]\left\lfloor[/itex][itex]\frac{x}{100}[/itex][itex]\right\rfloor[/itex]) - 9([itex]\left\lfloor[/itex][itex]\frac{x}{1000}[/itex][itex]\right\rfloor[/itex]) - 9([itex]\left\lfloor[/itex][itex]\frac{x}{10000}[/itex][itex]\right\rfloor[/itex]) ...

    For powers of ten where [itex]\frac{x}{10^{n}}[/itex] is less than 1, those items become 0.

    To use this formula quickly, you may wish to use decimals and do the floor function in your head.

    i.e.: Find the sum of the digits of 345854:
    345854 - 9(34585) - 9(3458) - 9(345) - 9(34) - 9(3) = 29
     
  7. May 12, 2013 #6
    Perfect! So it could be expressed as this:
    latex.codecogs.com/gif.latex?x%20-%209\sum_{n=1}^{\infty}{\left%20\lfloor%20\frac{x}{10^n}%20\right%20\rfloor}
     
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